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Question Number 39029 by MJS last updated on 01/Jul/18

$$\int \sqrt[3]{3-5\sqrt{1-\frac{1}{x}}}dx=?$$$$\int \frac{1}{\sqrt[3]{3-5\sqrt{1-\frac{1}{x}}}}dx=?$$

Answered by MJS last updated on 02/Jul/18

$$\int \sqrt[3]{3-5\sqrt{1-\frac{1}{x}}}dx=$$$$[t=-3+5\sqrt{1-\frac{1}{x}}\to dx=\frac{2}{5}{x}^2\sqrt{1-\frac{1}{x}}dt]$$$$=-50\int \frac{(t+3)\sqrt[3]{t}}{{(t-2)}^2{(t+8)}^2}dt=$$$$[u=\sqrt[3]{t}\to dt=3\sqrt[3]{u^2}du]$$$$=-150\int \frac{u^3(u^3+3)}{{(u^3-2)}^2{(u^3+8)}^2}du=$$$$\left[\begin{array}{c}{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\\ \int \frac{p}{q}=\frac{p_1}{q_1}+\int \frac{p_2}{q_2}\\ q_1=\gcd (q,q^{\prime });q_2=\frac{q}{q_1}\\ \mathrm{we}\mathrm{get}{p}_1,p_2\mathrm{by}\mathrm{matching}\mathrm{constants}\mathrm{of}\\ \frac{p}{q}=\frac{d}{du}\left(\frac{p_1}{q_1}\right)+\frac{p_2}{q_2}\\ p=u^3(u^3+3);q={(u^3-2)}^2{(u^3+8)}^2\\ q^{\prime }=6u^2(u^3-2)(u^3+8)(2u^3+6)\\ q_1=q_2=(u^3-2)(u^3+8)\\ p_1=-\frac{u}{6};p_2=\frac{1}{6}\end{array}\right]$$$$=25\frac{u}{(u^3-2)(u^3+8)}-25\int \frac{du}{(u^3-2)(u^3+8)}$$$$$$$$\int \frac{du}{(u^3-2)(u^3+8)}=\int \frac{du}{(u^3-2)(u^2-2u+4)(u+2)}=$$$$=\frac{1}{10}\int \frac{du}{u^3-2}+\frac{1}{120}\int \frac{u-4}{u^2-2u+4}du-\frac{1}{120}\int \frac{du}{u+2}$$$$$$$$\int \frac{du}{u+2}=\ln \mid u+2\mid$$$$\int \frac{u-4}{u^2-2u+4}du=\frac{1}{2}\int \frac{2u-2}{u^2-2u+4}du-3\int \frac{du}{u^2-2u+4}=$$$$[\int \frac{ax+b}{{ax}^2+bx+c}dx=\ln \mid {ax}^2+bx+c\mid ]$$$$[\int \frac{dx}{{ax}^2+bx+c}=\frac{2}{\sqrt{4ac-b^2}}\arctan \frac{2ax+b}{\sqrt{4ac-b^2}}]$$$$\frac{1}{2}\ln \mid u^2-2u+4\mid -\sqrt{3}\arctan \frac{\sqrt{3}}{3}(u-1)$$$$\int \frac{du}{u^3-2}=\int \frac{du}{(u-\sqrt[3]{2})(u^2+\sqrt[3]{2}u+\sqrt[3]{4})}=$$$$=\frac{\sqrt[3]{2}}{6}(\int \frac{du}{u-\sqrt[3]{2}}-\int \frac{u+\sqrt[3]{4}}{u^2+\sqrt[3]{2}u+\sqrt[3]{4}}du)=$$$$=\frac{\sqrt[3]{2}}{6}(\int \frac{du}{u-\sqrt[3]{2}}-\frac{1}{2}\int \frac{2u+\sqrt[3]{2}}{u^2+\sqrt[3]{2}u+\sqrt[3]{4}}-\frac{3\sqrt[3]{2}}{2}\int \frac{du}{u^2+\sqrt[3]{2}u+\sqrt[3]{4}})=$$$$=\frac{\sqrt[3]{2}}{6}(\ln \mid u-\sqrt[3]{2}\mid -\frac{1}{2}\ln \mid u^2+\sqrt[3]{2}u+\sqrt[3]{4}\mid -\sqrt{3}\arctan \frac{\sqrt{3}}{3}(\sqrt[3]{4}u+1))$$$$$$$$=-\frac{5\sqrt[3]{2}}{12}(\ln \mid u-\sqrt[3]{2}\mid -\frac{1}{2}\ln \mid u^2+\sqrt[3]{2}u+\sqrt[3]{4}\mid -\sqrt{3}\arctan \frac{\sqrt{3}}{3}(\sqrt[3]{4}u+1))-$$$$-\frac{5}{48}\ln \mid u^2-2u+4\mid -\frac{5\sqrt{3}}{24}\arctan \frac{\sqrt{3}}{3}(u-1)+$$$$+\frac{5}{24}\ln \mid u+2\mid +25\frac{u}{(u^3-2)(u^3+8)}+C$$$$u=\sqrt[3]{t}$$$$t=-3+5\sqrt{1-\frac{1}{x}}$$$$\mathrm{too}\mathrm{lazy}\mathrm{to}\mathrm{insert}\mathrm{these}...\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{solved}$$

Commented by maxmathsup by imad last updated on 02/Jul/18

$$thankyousirMjsyouhaveafastmemoryofcalculus...$$

Answered by MJS last updated on 02/Jul/18

$$...\mathrm{in}\mathrm{the}{2}^{\mathrm{nd}}\mathrm{case}\mathrm{the}\mathrm{same}\mathrm{substitutions}$$$$t=-3+5\sqrt{1-\frac{1}{x}}\mathrm{and}u=\sqrt[3]{t}\mathrm{lead}\mathrm{to}$$$$150\int \frac{(u^3+3)u}{{(u^3-2)}^2{(u^3+8)}^2}du\mathrm{and}\mathrm{this}\mathrm{can}\mathrm{be}\mathrm{solved}$$$$\mathrm{in}\mathrm{the}\mathrm{same}\mathrm{way}\mathrm{as}\mathrm{the}{1}^{\mathrm{st}}\mathrm{one}.$$

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